Let $K$ be a number field. If $X$ is a curve over $K$ with good reduction at a place $v$ of $K$, then the Jacobian of $X$ also has good reduction at $v$. This follows from the functoriality of the Jacobian.

The converse is not true, but I don't know of any examples.

Can one provide an example for all number fields $K$?

If not, I would also be pleased with just a counterexample for some place $v$ of some number field $K$

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    $\begingroup$ Hi, this is a nice question. Perhaps old work of Oda is relevant; see also mathoverflow.net/questions/69462/… Oda gave a necessary and sufficient criterion for good reduction of the curve in terms on the Galois action on the fundamental group of the curve. The Tate module of the Jacobian is the abelianised fundamental group. So one wants non-trivial action of inertia on $\pi_1$ but trivial on $\pi_1^{ab}$. see page 16 of igitur-archive.library.uu.nl/math/2006-1207-203237/… $\endgroup$ – SGP Mar 22 '12 at 13:29
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    $\begingroup$ As was pointed out by Emerton in answer to a previous question mathoverflow.net/questions/81501/…, if you take an elliptic curve $E$ with good reduction over $K_v$, then any $E$-torsor $P$ which has no $K_v$-rational point has bad reduction, still $\operatorname{Jac}(P) =E$ has good reduction. $\endgroup$ – François Brunault Mar 22 '12 at 16:39
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    $\begingroup$ No-one has made the following comment yet, so I will: if $X$ is a plane conic then $X$ typically has bad reduction at some non-zero set of primes, but its Jacobian is zero and hence has good reduction everywhere. $\endgroup$ – Kevin Buzzard Mar 22 '12 at 21:07

Let $E,E'$ be elliptic curves over the residue field. Then $E \times E'$ has good reduction but is not a Jacobian. However, any principally polarized abelian surface that reduces to $E \times E'$ but is not a product of elliptic curves is a Jacobian of some genus-$2$ curve $C$ that cannot have good reduction.

Explicitly (when the residue characteristic is odd), $C$ can have the form $y^2 = P(x)$ where $P$ is a sextic with roots $x_1,x_2,x_3$, $1/x'_1,1/x'_2,1/x'_3$ such that each $x_i,x'_i$ reduces to zero and each $x_i-x_j$ and $x'_i-x'_j$ ($i\neq j$) has valuation $1$. If I remember right, the Jacobian reduces to $E \times E'$ where $E: Y^2 = (X-\bar x_1) (X-\bar x_2) (X-\bar x_3)$ and $E': Y^2 = (X-\bar x'_1) (X-\bar x'_2) (X-\bar x'_3)$.

As I recall I learned this construction from Joe harris.

  • $\begingroup$ One thing Noam's example shows is that the Torelli map from $M_g$ (not compactified) to $A_g$ (again not compactified) is not a closed map, although it is a locally closed immersion (by the Torelli theorem). $\endgroup$ – Jason Starr Mar 23 '12 at 12:37

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